Relativistic Energy-Momentum Relation

According to Equations (3.162), (3.165), and (3.174), a particle of rest mass $m_0$, moving at velocity ${\bf v}$, has a relativistic momentum

$${\bf p} = \frac{m_0\,{\bf v}}{\sqrt{1-v^2/c^2}},$$ (3.176)

and a relativistic energy

$$E = \frac{m_0\,c^2}{\sqrt{1-v^2/c^2}}.$$ (3.177)

Thus,

$$\frac{E^2}{c^2}- \vert{\bf p}\vert^2 = \frac{m_0^2\,c^2}{1-v^2/c^2} - \frac{m_0^{\,2}\,c^2\,(v^2/c^2)}{1-v^2/c^2}= m_0^{\,2}\,c^2,$$ (3.178)

which leads to the relativistic energy-momentum relation,

$$\frac{E^2}{c^2} - \vert{\bf p}\vert^2 = m_0^{\,2}\,c^2.$$ (3.179)

Now, given that the rest mass is independent of the particle's motion (i.e., it is the same in all inertial frames of reference), we deduce that $E^2/c^2-\vert{\bf p}\vert^2$ takes the same value in all inertial frames of reference.